Question

How much heat, in kilojoules, is evolved in the complete combustion of (a) \(1.325 \mathrm{g} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm} ;\) (b) \(28.4 \mathrm{L} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(\mathrm{STP} ;(\mathrm{c})\) \(12.6 \mathrm{LC}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(23.6^{\circ} \mathrm{C}\) and \(738 \mathrm{mmHg} ?\) Assume that the enthalpy change for the reaction does not change significantly with temperature or pressure. The complete combustion of butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) is represented by the equation $$\begin{array}{r} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ}=-2877 \mathrm{kJ} \end{array}$$

Short answer

The heat evolved for the combustion of \(1.325g \) of \(C_{4}H_{10}\) is \(-65.1kJ\), for \(28.4L\) of \(C_{4}H_{10}\) at \(STP\) is \(-3407.3kJ\), and for \(12.6L\) of \(C_{4}H_{10}\) at \(23.6^{\circ}C\) and \(738mmHg\) is \(-1534.6kJ\).

Step-by-step solution

Mole Calculation

Calculate the number of moles of \(C_{4}H_{10}\) using the formula: \(n=\frac{m}{MW}\), where \(m=1.325g\) is the mass and \(MW=58.122g/mol\) is the molecular weight of butane.

Heat Calculation (Part 1)

Calculate the heat evolved in the combustion using the formula: \( q = n \cdot ΔH^{\circ}\), where \(n\) is the number of moles calculated and \(-2877 kJ/mol\) is enthalpy change. The sign of the heat evolved is negative because it's an exothermic reaction.

Mole Calculation (Part 2)

For the second condition, calculate the number of moles from the volume using the formula: \(n=\frac{V}{V_{m}}\), where \(V=28.4L\) is the volume and \(V_{m}=22.414L/mol\) at standard conditions.

Heat Calculation (Part 2)

Repeat the heat calculation with the new moles calculated.

Mole Calculation (Part 3)

For the third condition, calculate volume at standard conditions using the Ideal Gas Law. Then, calculate the mole from the volume at the standard condition using the same formula from Step 3.

Heat Calculation (Part 3)

Again, repeat the heat computation with the calculated moles.

Key concepts

Enthalpy Change

In thermochemistry, enthalpy change is a key concept that relates to the heat change occurring in a chemical reaction. It's symbolized by \( \Delta H \) and indicates the amount of energy absorbed or released to the surroundings during a reaction. If \( \Delta H \) is negative, the reaction is exothermic, meaning it releases heat. Conversely, if it's positive, the reaction is endothermic, absorbing heat from its surroundings.

Enthalpy change for a complete combustion reaction, like that of butane \( (\text{C}_4\text{H}_{10}) \), is typically pre-calculated and given, as it is in this problem: \( \Delta H^{\circ} = -2877 \text{ kJ/mol} \). This indicates a large amount of energy is released when butane combusts completely in the presence of oxygen, forming carbon dioxide \((\text{CO}_2)\) and water \((\text{H}_2\text{O})\).

To calculate the heat evolved during combustion, we use the formula:

• \( q = n \cdot \Delta H^{\circ} \)

where \( q \) is the heat evolved or absorbed, and \( n \) is the number of moles of butane involved. Understanding this concept helps in estimating the energy changes during reactions, which is crucial for both industrial applications and understanding energy efficiencies.

Combustion Reaction

A combustion reaction is a high-energy chemical reaction where a substance combines with oxygen to produce heat and, usually, light. It's a type of oxidation reaction and is essential for sustaining processes like automobile engine functioning and heating systems.

In the context of butane \( (\text{C}_4\text{H}_{10}) \), a common hydrocarbon, the combustion reaction involves butane completely reacting with oxygen \((\text{O}_2)\) to form carbon dioxide \((\text{CO}_2)\) and water \((\text{H}_2\text{O})\). The balanced equation is:
\[ \text{C}_4\text{H}_{10} + \frac{13}{2} \text{O}_2 \longrightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} \]

Complete combustion occurs when there is enough oxygen present to convert all the carbon and hydrogen in a hydrocarbon into \( \text{CO}_2 \) and \( \text{H}_2\text{O} \). This is key in calculating the enthalpy change, as variations in oxygen availability can lead to incomplete combustion, creating carbon monoxide instead.

It's important to remember that all hydrocarbons have different stoichiometric ratios and energy outputs, influencing their efficiency and applications. Understanding combustion reactions and their thermochemistry helps in optimizing fuel use and minimizing environmental impacts.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that provides a good approximation of how gases behave under various conditions of temperature and pressure. It's expressed as:\( PV = nRT \)

where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant (0.0821 L atm/mol K), and
• \( T \) is the temperature in Kelvin.

This law is particularly useful when dealing with gases under non-standard conditions, as no real gas behaves ideally under all circumstances.

In practice, for the exercise involving butane at different conditions, the Ideal Gas Law helps us calculate the volume of butane at Standard Temperature and Pressure (STP), where 1 mole of an ideal gas occupies 22.414 L. Using this law, conversions can be made from conditions not at STP, like for conditions given in c) \((23.6^\circ\text{C}, 738 \text{mmHg})\) to determine moles and thus calculate the heat evolved during the combustion of gases according to their environments.